Abstract

We show that an irreducible quasiprojective variety of dimension defined over an algebraically closed field with characteristic zero is an affine variety if and only if () = 0 and () = 0 for all , , where is any hypersurface with sufficiently large degree. A direct application is that an irreducible quasiprojective variety over is a Stein variety if it satisfies the two vanishing conditions. Here, all sheaves are algebraic.

1. Introduction

We work over an algebraically closed field with characteristic zero.

Affine varieties are important in algebraic geometry. J.-P. Serre introduced sheaf and cohomology techniques to algebraic geometry and discovered his well-known cohomology criterion ([1], [2, Chapter 2, Theorem 1.1]): a variety (or a Noetherian scheme) is an affine variety if and only if for all coherent sheaves on and all , . Goodman and Hartshorne proved that is an affine variety if and only if contains no complete curves and the dimension of the linear space is bounded for all coherent sheaves on [3]. Let be the completion of . In 1969, Goodman also proved that is affine if and only if after suitable blowing up of the closed subvariety on the boundary , the new boundary is a support of an ample divisor, where is the blowing up with center in ([4], [2, Chapter 2, Theorem 6.1]). For any quasiprojective variety , we may assume that the boundary is the support of an effective divisor with simple normal crossings by blowing up the closed subvariety in . is affine if is ample. So, if we can show the ampleness of , then is affine. There are two important criteria for ampleness according to Nakai-Moishezon and Kleiman ([5], [6, Chapter 1, Section 1.5]). Another sufficient condition is that if contains no complete curves and the linear system is base point free, then is affine [2, Chapter 2, Page 64]. Therefore, we can apply base point free theorem if we know the numerical condition of [6, Chapter 3, Page 75, Theorem 3.3]. Neeman proved that if is a quasicompact Zariski open subset of an affine scheme Spec, then is affine if and only if for all [7]. The significance of Neeman’s theorem is that it is not assumed that the ring is Noetherian.

In [8], we show that if a quasiprojective variety is Stein, for all , and has algebraically independent nonconstant regular functions, then is an affine variety. In this note, we give a new criterion for affineness.

Theorem 1. An irreducible quasiprojective variety of dimension is an affine variety if and only if for all , , and , where is any hypersurface with sufficiently large degree and .

By Cartan’s Theorem B, an analytic variety is a Stein variety if and only if for all analytic coherent sheaf on , for all . Since an algebraic affine variety over is a Stein variety [2, Page 232], we have the following application.

Corollary 2. An irreducible quasiprojective variety of dimension over is a Stein variety if for all , , and , where is any hypersurface with sufficiently large degree, , and all sheaves are algebraic.

Notice that in Theorem 1, the two vanishing conditions imply that any hypersurface section of is an affine variety. On the other hand, if every hypersurface section of is affine, may not be affine (Example 12).

If is an affine variety, then the ring is finitely generated [9, Page 20]. However, in our proof of Theorem 1, we do not directly check the finitely generated property of this ring, which is very hard in general. And the question of a quasiprojective variety to be affine is different from the behavior of the boundary divisor , in particular, the numerical condition of like nefness and finitely generated property of the graded ring

The reason is that

We will give two examples to demonstrate this difference in Section 3. One example (due to Zariski) is an affine surface such that the the corresponding graded ring is not finitely generated for an effective divisor . The other example [2, Page 232] is a surface such that

A necessary condition for the affineness of with dimension is that has plenty of nonconstant regular functions. More precisely, has algebraically independent nonconstant regular functions. This means that the corresponding effective boundary divisor must be big, that is,
for some positive number and . So, this surface is not affine but is finitely generated.

2. Proof of the Theorem

Recall our notation: is an open subset of a projective variety with dimension and is the effective boundary divisor with support . We may assume that has simple normal crossings by further blowing up suitable closed subvariety of .

In the following lemmas, is irreducible and satisfies and for all and , where is any hypersurface with sufficiently large degree in the ambient projective space containing .

Lemma 3. Let be a hypersurface in Theorem 1, , then satisfies the same vanishing conditions: and for all , where is any hypersurface section on with sufficiently large degree.

Proof. There is a short exact sequence,
where is considered as a Cartier divisor on . By the assumption, we have and for all . The corresponding long exact sequence gives for all . Similarly, for any hypersurfaces and , , from the short exact sequence
we have for all .

Lemma 4. If is a curve with for all hypersurface sections with sufficiently large degree, then is an affine curve.

Proof. First, we assume that is irreducible.If is complete, then by Riemann-Roch for singular curves [9, Page 298], we have
where is the arithmetic genus of and is the divisor with support in the set of smooth points of given by . Choose the hypersurface with sufficiently large degree such that
then . This is a contradiction. Therefore, is not a complete but an affine curve [2, Page 62].If is not irreducible, then is still an affine curve. Assume that has two irreducible components and , . Then, the dimension of is at most 0. So,
By Mayer-Vietoris sequence, for the ideal sheaf , we have
where the first and last terms vanish. This gives
Now and are affine curves. If has more than irreducible components, then by using mathematical induction, every irreducible component of is an affine curve. Thus is an affine curve.

By Lemmas 3 and 4 and mathematical induction, we may assume that every hypersurface section of is an affine variety.

Lemma 5. If is an irreducible surface with and for all for all hypersurface sections with sufficiently large degree, then is an affine surface.

Proof. By Lemma 3, any hypersurface section with sufficiently large degree on satisfies the same vanishing condition. So, is an affine curve by Lemma 4. Since is closed in , is not complete.Let be an irreducible curve on , then we may choose such that is irreducible [10] and contains more than two points [10]. Let and be two distinct points on , then there is a regular function on such that since is an affine curve. From the exact sequence
and , we have a surjective map from to . Lift from to , we have a regular function on such that it is not a constant on . By Goodman and Hartshorne’s theorem [3], is a quasiaffine variety. By Neeman's theorem [7], is an affine surface.

Let be an irreducible normal complete variety and let be a Cartier divisor on . If for all , then we define the -dimension to be . Otherwise, we define

If is not normal, we define , where is the normalization. From the definition, we see that if is an effective divisor, then , where is the dimension of . An effective divisor is defined to be big if .

Lemma 6. Let and be a hypersurface such that is irreducible, then is a big divisor on .

Proof. Let be the open irreducible hypersurface section, then it satisfies the same condition in Theorem 1 by the above lemmas. We may assume that is an affine variety by Lemmas 4 and 5 and inductive assumption. So, the closure in has algebraically independent nonconstant rational functions which are regular on . This implies that is a big divisor on , that is, .

Lemma 7. has no complete curves.

Proof. If has an irreducible complete curve , choose a hypersurface such that is irreducible [10] and intersects at more than 2 distinct points. Let , . By Lemmas 4 and 5, we may assume that any irreducible hypersurface section of dimension with sufficiently large degree is an affine variety. By inductive assumption, is an affine variety. So, there is a regular function such that . Since
we can lift to . So there is a regular function on such that is not a constant. This is not possible since is complete. The contradiction shows that has no complete curves.

By Lemma 4 and inductive assumption, if is not irreducible, the proof still works since is affine.

Lemma 8. For any irreducible curve on , there is a regular function on such that is not a constant on .

Proof. Let be an irreducible complete curve in containing . Then, is a finite set and a general hypersurface does not contain any point in . Let be a hypersurface away from such that intersects at more than two distinct points and is irreducible [10]. Then, is an affine variety and . Let and be two distinct points in , then there is a regular function on such that . Lift to , we find a regular function on such that and is not a constant.

Theorem 9. An irreducible quasiprojective variety is an affine variety if and only if and for all , , where is any hypersurface with sufficiently large degree.

Proof. By Serre’s criterion, one direction is trivial: if is affine, then it satisfies and for all and , where is any hypersurface.We assume that satisfies and for all .By Lemma 8 and Goodman and Hartshorne’s theorem [3], is a quasiaffine variety since for every irreducible curve on , there is a global regular function on such that is not a constant. By Neeman’s result [7], a quasiaffine variety is affine if and only if for all . So, is an affine variety.

If is not irreducible, then Theorem 9 still holds since the proof works by Lemma 4 and mathematical induction.

Corollary 10. An irreducible quasiprojective variety of dimension over is a Stein variety if for all , , , and , where is any hypersurface with sufficiently large degree, and all sheaves are algebraic.

3. Examples

Again is an irreducible open variety contained in a projective variety such that , where is an effective boundary divisor with support . In this section, we assume that the ground field is .

Example 11. There is an affine surface such that the graded ring
is not finitely generated. This example is according to Zariski [11, Pages 562–564].Let be a smooth curve of degree 3 in . Let be a divisor class cut out on by a curve of degree 4 in . There exist 12 distinct points on such that
for all positive integers . Let be a surface obtained by blowing up at these 12 points . Let be the strict transformation of (i.e., the closure of the inverse image of in ). Let be a line not passing through any point in these 12 points. Let be the strict transform of . Then, the complete linear system
has a fixed locus for all and
has no fixed components and is base point free. By Nakai-Moishenzon’s ampleness criterion [9, Chapter V, Section 1], the divisor
is ample. Hence, the complement is affine but the graded ring
is not finitely generated.

Example 12. A nonaffine surface such that the graded ring
is finitely generated.

Let be an elliptic curve and the unique nonsplit extension of by itself. Let and be the canonical section, then is not affine and [2, Page 232]. So,
is finitely generated.

Since the surface has no complete curves, any hypersurface section of is an affine curve. But is not affine since [2, Page 232]. It shows that if every hypersurface section of is affine, may not be affine.

The above two examples demonstrate that the affineness of and the finitely generated property of the graded ring
are different in nature. The reason is that

In fact, we have the following.

Lemma 13 (see [3]). Let be a scheme and be an effective Cartier divisor on . Let and be any coherent sheaf on , then, for every ,

So, we have

The direct limit is the quotient of the direct sum and its subring, so it is much “smaller” than direct sum [12, Chapter II, Section 10]. And even though is affine, the boundary divisor can be very bad. For example, may not be nef. It is easy to see this by blowing up at a point. Let be a line in , let be a point on . Let be the blow up of at . Let be the exceptional divisor and , where is the strict transform of and is a large positive integer. Then, [2, Chapter V, Corollary 3.7]. Therefore, is not nef.

Example 14. If a smooth threefold such that contains no complete curves, then for all and but is not affine.

Let be a smooth projective elliptic curve defined by , . Let be the elliptic surface defined by the same equation, then we have surjective morphism from to such that for every , the fiber . In [13], we proved that there is a rank 2 vector bundle on such that when restricted to , is the unique nonsplit extension of by , where is the morphism from to . We also proved that there is a divisor on such that when restricted to , is the canonical section of . Let , we have for all and . We know that this threefold contains no complete curves [13] and . So, is not affine.

Example 15.
A surface without complete curves such that but is not affine.

Remove a line from , then we have . Remove the origin from , let . Then, is not affine since the boundary is not connected [2, Chapter II, Section 3 and Section 6]. Blow up with center , let be the exceptional divisor and be the blowup. Let , where is the strict transformation of . Then by Iitaka’s result, on , and has no complete curves, but is not affine.

Lemma 16 (see [6]). Let be a surjective map between projective varieties, smooth, normal. Let be the geometric generic fiber of and assume that is connected. The following two statements are equivalent:(i) for all ;(ii) has rational singularities and for all .

Example 17. A smooth variety of dimension with for all and but is not affine.

Let be the smooth projective variety obtained by blowing up a point in . Let be the blowup. Let be a hyperplane not passing through . Let and . Then, [14, Chapter 2, Theorem 5.13]. Let be the exceptional divisor on , then . So, for all . By Lemma 16, we have for all .

Let , then . For the global sections on affine space , we have [9, Chapter III, Proposition 8.1, 8.5 and Chapter II, Proposition 5.1(d)]
for all .

It is obvious that is not affine since it contains a projective space .

Example 18. A threefold satisfies the following three conditions but is not affine:(1) contains no complete curves;(2)the boundary is connected;(3).

Let be a hyperplane in . Let be a line not contained in . Blow up along , let be the blowup. Define a divisor on such that , where is the exceptional divisor on . Let , then .

It is easy to see that the above three conditions are satisfied. is an open subset of and is a line, which is not of codimension 1. So, is not affine [2, 7].

Example 19. A quasiprojective variety with a surjective morphism such that is affine, a general fiber is affine and for all , but is not affine.

Let , be the varieties defined in Example 17. Then the fiber space satisfied the above requirements. is not affine because it has a projective space .